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MathRevolution
[GMAT math practice question]

Alice makes 2-digit codes using the 26 letters of the alphabet. Letters may be used repeatedly, and at least one vowel must be used. How many possible codes can she make?

A. 235
B. 256
C. 360
D. 625
E. 676

To be honest Y is kind of vowel too. I think that the Q should be modified, telling how many vowels exist, otherwise one can assume (and I did) in my calculations that there are 6 vowels https://simple.wikipedia.org/wiki/Vowel
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MathRevolution
[GMAT math practice question]

Alice makes 2-digit codes using the 26 letters of the alphabet. Letters may be used repeatedly, and at least one vowel must be used. How many possible codes can she make?

A. 235
B. 256
C. 360
D. 625
E. 676


Two ways...
First the better approach..
Total ways =26*25=650
Ways both are consonants = 21*20=420..
Atleast one vowel = 650-420=230

Second approach..
Ways both are vowels = 5*4=20
Ways one is consonant and one vowel = 21*5=105
These can be placed in two ways... CV or VC, so 105*2=210
Total = 20+210=230..

I suppose A is meant to be 230..
-------------------------------------------------------------------------------------------------------------

Shouldn't the answer be 235 because the question says that the alphabets can be repeated?

Yes, thanks.
Without repetition, it will be 230..
And with repetition, 5 cases get added.. as,ee,ii,oo,uu
So 230+5=235
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=>

The number of vowels is 5 (a,e,i,o,u). So, the number of codes is equal to the number of 2-digit codes with no restrictions minus the number of codes that contain no vowels. This is given by

26*26 – 21*21 = (26 + 21)(26 – 21) = 47*5 = 235.

Therefore, the answer is A.
Answer: A
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MathRevolution
[GMAT math practice question]

Alice makes 2-digit codes using the 26 letters of the alphabet. Letters may be used repeatedly, and at least one vowel must be used. How many possible codes can she make?

A. 235
B. 256
C. 360
D. 625
E. 676

Each code must have at least one vowel. Thus, we have 3 cases for creating the codes: 1) vowel-consonant, 2) consonant-vowel, and 3) vowel-vowel.

Case 1: vowel-consonant

We have 5 x 21 = 105 possible codes.

Case 2: consonant-vowel

We also have 21 x 5 = 105 possible codes.

Case 3: vowel-vowel

We have 5 x 5 = 25 possible codes.

Thus, the total number of possible codes is 105 + 105 + 25 = 235.

Answer: A
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5*26*2 = 260
(five possible vowels as the first digit times all possibilities of the second digit ; times two so vowels take the second digit instead)

260 - 5*5 = 235
(minus the combinations of both being vowels since 5*26*2 would repeat the same vowel to vowel combinations twice)

Is this simpler and does this make sense?
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